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UPRVUNL JE EE 2014 Official Paper

Option 3 : 19.6 A

ST 1: General Knowledge

5609

20 Questions
20 Marks
20 Mins

**Concept:**

**Transformer on No-load:**

When the transformer is on no-load, the primary input current is not wholly reactive. The primary input current under no-load conditions has to supply due to the following

- Iron losses in the core i.e. hysteresis loss and eddy current loss
- A very small amount of copper loss in primary

The no-load primary input current I_{0} is not at 90° behind supply voltage but lags it by an angle ϕ_{0} < 90°.

No-load input power (W_{0}) is given as,

W_{0} = V_{1 }I_{0 }cos ϕ_{0}

Where,

V_{1} is the primary supply voltage

I_{0} is no-load current

cos ϕ_{0} is no-load power factor

The primary no-load current I_{0} has two-component as shown in phasor,

Where,

I_{M} is a magnetizing component of no-load current.

I_{u} is an active or working or iron loss component of no-load current.

I_{M} = I_{0} sin ϕ_{0}

I_{u} = I_{0} cos ϕ_{0}

\(I_0= \sqrt{{I_M}^2+{I_u}^2}\)

**Calculation:**

Given,

I_{0} = 20 A

cos ϕ_{0} = 0.20 lagging

∴ ϕ_{0} = cos^{-1 }(0.20) = 78.46

From the above concept,

IM = I0 sin ϕ0 = 20 × sin (78.46) = 19.6 A

I0 = 20 A

cos ϕ0 = 0.20 lagging

Iu = I0 cos ϕ0 = 20 × 0.20 = 4

\(I_0= \sqrt{{I_M}^2+{I_u}^2}\)

\(20= \sqrt{{I_M}^2+{4}^2} \)

I_{M}^{2} = 400 - 16 = 384

I_{M} = 19.6 A